Prove $\exists x$ st $\|x\|=1$ and $\operatorname{dist}(x,Y)=1$ 
If $Y$ be a  finite-dimensional proper subspace of a normed linear space $X$, then show that there exists $x\in X$ st $\|x\|=1$ and $\operatorname{dist}(x,Y)=1$

For any $x \in X$ st $\|x\|=1$ we have $\operatorname{dist}(x,Y)\le 1$ since $0\in Y$, so we need to show there exists a point $x\in X$ with $\|x\|=1$ and $\operatorname{dist}(x,Y)\ge 1$ that will complete the proof.
 A: First show that for any $x\in X$, $\exists y_0\in Y:\|x-y_0\|=\text{dist}(x,Y)$. 
For this, let $a=\text{dist}(x,Y)$. So there is a sequence $\{y_n\}\in Y$ such that $\|x-y_n\|\le a+\frac{1}{n}$. So $y_n$ is a bounded sequence in $Y$. Let $\|y_n\|\le M$ (say), for all $n$. Hence the subset $A=\{y\in Y:\|y\| \le M\}$ is a closed and bounded subset of the finite dimensional subspace $Y$ and hence $A$ is compact. So $y_n$ has a convergent subsequence, say $y_{n_k}\to y_0$ in $A$. Now $\|x-y_0\|\ge \text{dist}(x,Y)=a$ and $\|x-y_0\|\le \|x-y_{n_k}\|+\|y_{n_k}-y_0\|\le a+ \frac{1}{n_k}+\|y_{n_k}-y_0\|$, for all $k$. Hence taking limit, $a\le \|x-y_0\|\le a$ i.e. $\|x-y_0\|=a=\text{dist}(x,Y)$.
For the remaining part, proceed as follows:
Let $v\in X\setminus Y$ and let $b=\text{dist}(v,Y)$. Clearly $b>0$ as $Y$ (being finite dimensional) is closed. By the first part, we have $\exists y_0\in Y$ such that $b=\|v-y_0\|$. So choose $x=c(v-y_0)$, where $c=\frac{1}{\|v-y_0\|}=\frac{1}{b}$. Clearly $\|x\|=1$ and for any $y\in Y$,
$$\|x-y\|=\|c(v-y_0)-y\|=c\|v-y_0-c^{-1}y\|=c\|v-y_1\|,$$
where $y_1=y_0+c^{-1}y\in Y$. Hence $\|x-y\|\ge c \cdot \text{dist}(v,Y)=cb=1$. Hence $\text{dist}(x,Y)\ge 1$ as required.
